In this expository article, we wish to investigate certain Diophantine equations arising from the Pythagorean theorem for integers and look at ways of investigating some of them. We also offer a proof of Fermat’s Last Theorem for the case . The contents in this document are neither new nor original; it is merely an attempt to test my own understanding of the material.

The equation of the form

The Pythagoras theorem tells us a fact about right angled triangles: that if two side lengths of a right angled triangle (other than that of the hypotenuse) be and and the hypotenuse be of length , then . This motivates us to find all positive integral solutions of the form .

More specifically, we shall investigate equations of the form where .

Such triples are known as primitive Pythagorean triples.

Two lemmas are in order.

Lemma 1: In a Pythagorean triple,one and only one of and is even.

Proof:Note that by the definition of a primitive Pythagorean triplet. If both are even, . If both are odd, , a contradiction as any square is either or .So one of them is even, the other is odd.

Lemma 2: If and , then there exist positive integers such that .

: The proof is left to the reader.However, looking at the prime representation of and is helpful.

An Important Theorem

Theorem 1:The solutions to the equation where , , is given by

for some naturals and .

Proof: . Note that

If not, let . Then and .

Thus divides and . As , also divides which forces which is not true as their gcd is .

Using lemma 2,

and for some positive integers .

So, and , . Note that means which in turn implies that .Further, forces to be 1 as well.

QED.

Two More Theorems

Theorem 2: There is no solution to in positive integers .

The proof of this theorem will need some work. We will see repeated application of theorem 1 and the two lemmas preceding it.

Proof :We prove this by contradiction.

Suppose that the equation has a solution .

Further, let . So there exist natural numbers such that so that , .

That implies for some in naturals.

We now have such that .

So we now have a Pythagorean triple and so we have (assuming is even) for some positive naturals with .

and are odd by lemma 1.

If is odd, , a contradiction ( note that if t is odd, s is even). So, is even.

Let . Then such that .

Consider the equation (note that implies that is 1. So, there exist natural numbers such that .

Once again, and for some natural numbers using the previous arguments.

So, . Note that this makes a solution set to our equation.

i.e for each , there is a smaller which is part of the solution of the equation. being finite, at some stage, a contradiction.This is the Fermat’s method of infinite descent.

So the equation has no solution in natural numbers.

Note that this implies that that there is no solution to the equation , the Fermat’s Last Theorem for the exponent .

Theorem 3: There is no solution to the equation in natural numbers.

Proof: Imitating theorem 1’s proof, we consider that case when for coprime .

So now there are two cases. If is odd,for some natural numbers , where and and .

Multiplying the first two equations, we get i.e. ,

: the Fermat’s Method of infinite descent shows that there is a contradiction!

We are left with the case is even.

Then for some natural numbers , where and .

So, i.e and for some . So, for some .

We have . So, is a Pythagorean triple. So,

That allows us to conclude that are perfect squares. , . So i.e. is a solution of the original equation.

Note that and this produces a contradiction by Fermat’s method of infinite descent.